Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $k = \dfrac{-5a^2 - 15a - 10}{-5a^3 + 15a^2 + 20a} \times \dfrac{a^2 - 4a}{a - 6} $
Solution: First factor out any common factors. $k = \dfrac{-5(a^2 + 3a + 2)}{-5a(a^2 - 3a - 4)} \times \dfrac{a(a - 4)}{a - 6} $ Then factor the quadratic expressions. $k = \dfrac {-5(a + 1)(a + 2)} {-5a(a + 1)(a - 4)} \times \dfrac {a(a - 4)} {a - 6} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac { -5(a + 1)(a + 2) \times a(a - 4)} { -5a(a + 1)(a - 4) \times (a - 6)} $ $k = \dfrac {-5a(a + 1)(a + 2)(a - 4)} {-5a(a + 1)(a - 4)(a - 6)} $ Notice that $(a + 1)$ and $(a - 4)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {-5a\cancel{(a + 1)}(a + 2)(a - 4)} {-5a\cancel{(a + 1)}(a - 4)(a - 6)} $ We are dividing by $a + 1$ , so $a + 1 \neq 0$ Therefore, $a \neq -1$ $k = \dfrac {-5a\cancel{(a + 1)}(a + 2)\cancel{(a - 4)}} {-5a\cancel{(a + 1)}\cancel{(a - 4)}(a - 6)} $ We are dividing by $a - 4$ , so $a - 4 \neq 0$ Therefore, $a \neq 4$ $k = \dfrac {-5a(a + 2)} {-5a(a - 6)} $ $ k = \dfrac{a + 2}{a - 6}; a \neq -1; a \neq 4 $